See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: 2-Phenylpropene is C6H5-C(CH3)=CH2, i.e., an alkene with a phenyl group and a methyl group on the same carbon (C2), with a terminal double bond (C1=C2). The target molecule has the structure Ph-C(=CH2)-CH3. Step 1 - Identify the target: 2-Phenylpropene = Ph-C(CH3)=CH2. This is a trisubstituted/disubstituted alkene where the double bond carbon bearing the phenyl group also bears a methyl group. Step 2 - Retrosynthetic analysis: Working backward via dehydration, the precursor alcohol would be Ph-C(CH3)(OH)-CH3, i.e., 2-phenyl-2-propanol (a tertiary alcohol). Dehydration of this tertiary alcohol with H2SO4/heat would give 2-phenylpropene via E1 elimination (Zaitsev product, and the only possible alkene from a symmetrical tertiary alcohol here gives Ph-C(=CH2)-CH3). Step 3 - Synthesis of 2-phenyl-2-propanol: React phenylmagnesium bromide (PhMgBr, made from bromobenzene + Mg in diethyl ether) with acetone [(CH3)2C=O]. The Grignard reagent (PhMgBr) adds to acetone to give, after H3O+ workup, Ph-C(CH3)2-OH = 2-phenyl-2-propanol. Step 4 - Final step: Treatment of 2-phenyl-2-propanol with H2SO4 and heat causes dehydration to yield 2-phenylpropene (Ph-C(CH3)=CH2). Step 5 - Why other options fail: (a) Friedel-Crafts alkylation of benzene with 2-chloropropene and AlCl3: The allylic carbocation from 2-chloropropene would rearrange, and the alkylation product would not cleanly give 2-phenylpropene; this is not a reliable synthesis. (b) Benzaldehyde + EtMgBr gives Ph-CH(OH)-CH2CH3 (a secondary alcohol, 1-phenyl-1-propanol). Dehydration would give Ph-CH=CH-CH3 (1-phenylpropene / beta-methylstyrene), NOT 2-phenylpropene. (c) PhMgBr + propanal (CH3CH2CHO) gives Ph-CH(OH)-CH2CH3 (secondary alcohol). Dehydration gives Ph-CH=CH-CH3, not 2-phenylpropene. (d) PhMgBr + acetone gives Ph-C(CH3)2-OH. Dehydration gives Ph-C(CH3)=CH2 = 2-phenylpropene. Correct. Therefore, the correct answer is D.